2

$b^{n+1}$ - $a^{n+1}$ > $(n+1)*a^n*(b-a)$ if b > a > 0

$b^{n+1}$ > $[(n+1)*(b-a) + a]*a^n$

let $b=1+1/n; a=1+1/(n+1)$ then

$(1+1/n)^{n+1}$ > $[(n+1)*(1/n-1/(n+1)) + 1 + 1/(n+1)]*(1+1/(n+1))^n$

$(1+1/n)^{n+1}$ > $[1+1/n + 1/(n+1)]*(1+1/(n+1))^n$

$(1+1/n)^{n}$ > $[((1+1/n + 1/(n+1))/(1+1/n)]*(1+1/(n+1))^n$

$(1+1/n)^{n}$ > $[((1+1/n + 1/(n+1))/(1+1/n)^2]*[(1+1/n)*(1+1/(n+1))^{n}]$

$(1+1/(n-1))*(1+1/n)^{n}$ > $[((1+1/(n-1))*(1+1/n + 1/(n+1)))/((1+1/n)^2*(1+1/(n+1))]*[(1+1/n)*(1+1/(n+1))^{n+1}]$

it can be proved that $[((1+1/(n-1))*(1+1/n + 1/(n+1)))/((1+1/n)^2*(1+1/(n+1))] > 1$ so

$(1+1/(n-1))*(1+1/n)^{n}$ > $(1+1/n)*(1+1/(n+1))^{n+1}$

i.e $(1+1/n)^{n}/(1-1/n)$ > $(1+1/(n+1))^{n+1}/(1-1/(n+1))$

$(1+1/n)^{n}/(1-1/n)$ is a decreasing sequence

with starting case n=11 verified less than 3, so all terms in the sequence are less than 3

Q.E.D

Tx1 Nv
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  • this is hard to read. Please edit for clarity and to show your efforts. – lulu May 31 '23 at 22:59
  • I input n=11 and verified the numbers are valid for starting case. I'm lost at how to continue with induction step. I tried both subtraction and division between n and n+1 terms it lead to nowhere. – Tx1 Nv May 31 '23 at 23:09
  • Is the textbook itself requiring you to use induction or is that more your own goal? Because you can get this done in like 2 or 3 steps with limits lol. Of course induction is possible too but it'll require a bit more time to explain than I have right now, so if you're really wanting something based explicitly on induction... then watch this space :D – H. sapiens rex Jun 01 '23 at 00:02
  • Thanks for your input. The textbook didn't give hint or solution to the problem. From the wording of the problem, seems it's doable with induction. Also it's at the beginning of the book before limit/supremum had been introduced. – Tx1 Nv Jun 01 '23 at 00:26
  • I haven't checked this one, but often inequalities involving the sequence $(1+\dfrac1n)^n$ can be handled with Bernoulli's inequality: $$(1+x)^n>1+nx$$ for all $x>0$. – Jyrki Lahtonen Jun 01 '23 at 03:00

2 Answers2

1

Taking logs makes the problem easier. You want to show that $$n\ln\Big(1+\dfrac{1}{n}\Big)<\ln3 + \ln\Big(1-\dfrac{1}{n}\Big).$$

Now $\ln(1+x)<x$ and from this, substituting $x=\frac{1-y}{y}$, where $y<1$, $$-\ln y=\ln\Big(\dfrac{1}{y}\Big)<\dfrac{1}{y}-1.$$ Then $$n\ln\Big(1+\dfrac{1}{n}\Big)- \ln\Big(1-\dfrac{1}{n}\Big) <n\cdot \dfrac{1}{n}+\dfrac{1}{1-\dfrac{1}{n}}-1 =\dfrac{n}{n-1}.$$ Unfortunately, this is only less than $\ln 3$ for $n=12$ and above, so you will need to do the $n=11$ case by hand, using the numerical values given.

You could also use this method to prove the result by induction: then you would need to show that $$\dfrac{\Big(1+\dfrac{1}{n+1}\Big)^{n+1}}{\Big(1+\dfrac{1}{n}\Big)^{n}}<\dfrac{1-\dfrac{1}{n+1}}{1-\dfrac{1}{n}}$$ which can be written as $$\Big(1-\dfrac{1}{n^2}\Big)\Big(1-\dfrac{1}{(n+1)^2}\Big)^{n+1}<1-\dfrac{1}{n+1}.$$ Taking logs, we require: $$\ln\Big(1-\dfrac{1}{n^2}\Big)+(n+1)\ln\Big(1-\dfrac{1}{(n+1)^2}\Big)-\ln\Big(1-\dfrac{1}{n+1}\Big)<0.$$ and using $\ln\Big(1-\dfrac{1}{n^2}\Big)<-\dfrac{1}{n^2}$, $(n+1)\ln\Big(1-\dfrac{1}{(n+1)^2}\Big)<-\dfrac{1}{n+1}$ and $-\ln\Big(1-\dfrac{1}{n+1}\Big)<\dfrac{1}{1-\dfrac{1}{n+1}}-1$ gives, for the left hand side, $-\dfrac{1}{n^2(n+1)}$ showing the required inequality.

mcd
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For $x > 1$ let

\begin{equation}f \left(x\right) = x \ln \left(1+\frac{1}{x}\right)-\ln \left(1-\frac{1}{x}\right)\end{equation}

Its derivative is

\begin{equation}{f'} \left(x\right) = \ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}-\frac{1}{x \left(x-1\right)}\end{equation}

hence

\begin{equation}{f'} \left(x\right) \leqslant \frac{1}{x}-\frac{1}{x+1}-\frac{1}{x \left(x-1\right)} =-\frac{1}{x \left({x}^{2}-1\right)} < 0\end{equation}

Thus $ f$ decreases in $ \left(1 , \infty \right)$. For $ x \geqslant 11$ we have

\begin{equation}f \left(x\right) \leqslant f \left(11\right) = 10 \ln \left(\frac{12}{11}\right)+\ln \left(\frac{12}{10}\right) = \ln \left({{\left(\frac{12}{11}\right)}^{10}}\times{1.2}\right)\end{equation}

It follows that for such $ x \geqslant 11$,

\begin{equation}{\left(1+\frac{1}{x}\right)}^{x} \leqslant {{\left(1-\frac{1}{x}\right)}\times{2.3872}}\times{1.2} \leqslant {\left(1-\frac{1}{x}\right)}\times{3}\end{equation}

Gribouillis
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